Find the value of $ \lim_{n\rightarrow\infty}\sqrt[n]{\int_{1}^{3}x^{n/x} dx}$.
My attempt:
$$ L = (\lim_{x \to \infty}\int_{1}^{3} x^{\frac {n}{x}}\mathrm{d}x)^\frac {1} {n} \Longrightarrow \log L = \frac {\lim_{x \to \infty}\int_{1}^{3} x^{\frac{n}{x}}\mathrm{d}x} {n}$$
Using L'H$\hat{\mathrm{o}}$pital's rule,
$ \Longrightarrow \log L = \frac {\lim_{x \to \infty}\frac {\int_{1}^{3} \log x x^{\frac{n}{x}} \mathrm{d}x} {x}} {\lim_{x \to \infty}\int_{1}^{3} x^{\frac{n}{x}}\mathrm{d}x} = \frac {L \log 3} {3} $
Comparing $\frac {\log 3}{3} = \frac {\log L}{L}$,
$ L = 3$.
But the given answer is $ e^{1/e}$.